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A normal mode of a linearized mechanical system is a motion in which every generalized displacement oscillates at a single common angular frequency, with fixed relative amplitudes and phases. In coordinates $q_j$, an undamped oscillatory mode may be writtenwhere the amplitude vector $(A_1,\dots ,A_n)$ is fixed; negative components describe relative phase $\pi$. An arbitrary small motion is a linear combination of the normal modes when the mode vectors form a complete basis.

Regard $K$ as a vector space over $F$. Since $K$ is finite, a maximal linearly independent list has finite length $d$ and is a basis. Every element is uniquely a linear combination of its $d$ basis vectors, with $q$ choices for each coefficient. Hence $\boxed{|K|=q^d}$, with $d=[K:F]$. This includes $d=1$ when the extension is trivial.

A linear combination of independent normal random variables is normal, with variances adding when the coefficients are $1,-1$. Applying the joint central limit approximation from (i) givesEquivalently the difference minus $p-q$, divided by the displayed standard deviation, converges in distribution to a standard normal under the same regular large-sample regime.

In either exterior region of the finite square well, the time-independent Schrödinger equation reduces to ${\psi }^{\prime \prime }=k^2\psi$, since $E<0$. Its general solution is a linear combination of $e^{kx}$ and $e^{-kx}$. On the left, $e^{-kx}$ grows as $x\to -\infty$ and cannot be a square-integrable function, so its coefficient must be zero. On the right, the excluded growing term is $e^{kx}$. Hence the surviving tails are $A{e}^{kx}$ for $x\le -a$ and $D{e}^{-kx}$ for $x\ge a$.

Inside the well, $E+V_0>0$ and the equation becomes ${\psi }^{\prime \prime }=-{\kappa }^2\psi$, whose general solution is $B\cos \kappa x+C\sin \kappa x$. ThusThe coefficients are initially independent; wavefunction matching at a finite potential step imposes the remaining relations. Here the paper's $k$ is the exterior decay constant and $\kappa$ is the interior oscillation constant.

The position operators and momentum operators are self-adjoint operators. Using conjugate linearity of the adjoint operator, $(cA+dB{)}^{\ast }=\overline{c}{A}^{\ast }+\overline{d}{B}^{\ast }$, givesWrite $D=\sqrt{2m\omega \hslash }$. Bilinearity and antisymmetry of the commutator, together with the canonical commutation relations, give $[P_j,X_k]=-i\hslash {\delta }_{jk}I$. HenceThe terms involving two position operators or two momentum operators vanish by the canonical commutation relations. Finally, reversing products under the adjoint operator, $(AB{)}^{\ast }=B^{\ast }{A}^{\ast }$, gives $[a_j,a_k{]}^{\ast }=-[a_j^{\ast },a_k^{\ast }]$. ThereforeHere $I$ is the identity operator. These are the relations for two independent pairs of creation and annihilation operators. The operator algebra in this question is understood on a common invariant operator domain, such as finite linear combinations of the number states constructed in part (c).

Use part (a) to put the four projective points atA quadratic polynomial vanishing at the first three has no square terms, and vanishing at the fourth givesThese form a two-dimensional vector space. Choose the two membersTheir coefficient triples are independent and both satisfy the relation. Their symmetric bilinear form matrices are of the formFor $(A,B,C)=(1,1,-2)$ and $(1,-2,1)$, both determinants are $-1/2$, so both selected conics are nonsingular. Any coefficient triple with zero sum is uniquely a linear combination of these two. Hence every conic through the four points has matrixUndoing the projective transformation carries the entire family back to the original four points, preserving nonsingularity. This is a pencil of conics through four points.

For $L_i={ϵ}_{ijk}{q}_j{p}_k$, the canonical Poisson brackets give $\{L_i,L_j\}={ϵ}_{ijk}{L}_k$. Consequently projections onto fixed axes $n,m$ obeyIf the two original projections are conserved and the axes are linearly independent, the preceding Jacobi identity argument makes the projection along $n\times m$ conserved as well. The vectors $n,m,n\times m$ form a basis of three-dimensional space, so every fixed-axis projection is a linear combination of these conserved quantities. Angular momentum is therefore conserved along every axis.

The velocity coupling changes the momenta and the undifferentiated field terms. Variation with respect to ${\varphi }_2$ gives the equation for ${\varphi }_1$:so the two Euler-Lagrange field equations areThey describe damped and antidamped fields from an antisymmetric velocity coupling. With the convention ${\varphi }_j=A_j{e}^{i(k_{j}x-{\omega }_{j}t)}$, their characteristic equations areThus the dispersion relations and modes areFor each spatial Fourier mode, the general solution is the linear combination of its two temporal branches; spatial superposition gives the general field. Take real parts or suitable complex conjugate pairs to obtain the required real fields. At the repeated-root threshold, the independent second temporal solution contains an extra factor $t$, as detailed next.

Add the one-particle energy term $H_0={ϵ}_1{n}_1+{ϵ}_2{n}_2$. It is diagonal in the same Fock state basis as the contact interaction between two ring modes. For $N=2$, the ordered basis $(|2,0⟩,|1,1⟩,|0,2⟩)$ givesFor $g>0$, the middle eigenvalue exceeds the mean of the two endpoint eigenvalues by $g/L$, so it cannot be the smallest. The ground state is therefore $|2,0⟩$ if ${ϵ}_1<{ϵ}_2$, and $|0,2⟩$ if ${ϵ}_2<{ϵ}_1$, withIf ${ϵ}_1={ϵ}_2$, every normalized linear combination of the two endpoint Fock states is a ground state. The same endpoint argument works for general fixed $N$, since the interaction contribution is a concave quadratic function of the mode occupation number.

The rank-nullity theorem states that a linear map $T:V\to W$ with finite-dimensional $V$ satisfiesChoose a basis $k_1,\dots ,k_r$ of $\ker T$ and use basis extension to obtain a basis $k_1,\dots ,k_r,u_1,\dots ,u_s$ of $V$. We prove that $T{u}_1,\dots ,T{u}_s$ is a basis of $\mathrm{im}T$.

Every $v\in V$ is a linear combination of the extended basis. Applying $T$ kills the $k_i$ terms, so $Tv$ is a linear combination of $T{u}_1,\dots ,T{u}_s$. This proves the linear span assertion. For linear independence, if ${\sum }_j{b}_{j}T{u}_j=0$, then ${\sum }_j{b}_j{u}_j$ belongs to the kernel of a linear map. Hence ${\sum }_j{b}_j{u}_j={\sum }_i{a}_i{k}_i$ for some scalars $a_i$. The linear independence of the extended basis forces every $a_i$ and $b_j$ to vanish. Consequently $\mathrm{rank}T=s$ and $\mathrm{nullity}T=r$, while $\dim V=r+s$, proving the theorem.

The image of a linear map and kernel of a linear map areA linear map sends zero to zero, since $T(0)=T(0+0)=T(0)+T(0)$. Thus both displayed sets contain zero. If $w_1=T(v_1)$ and $w_2=T(v_2)$ belong to the image of a linear map, then for any real scalars $a,b$,If $v_1,v_2$ belong to the kernel of a linear map, thenso $a{v}_1+b{v}_2\in \ker T$. Containing zero and being closed under linear combinations is the vector subspace criterion. Hence the image of a linear map is a vector subspace of $W$, and the kernel of a linear map is a vector subspace of $V$.

The rank-nullity theorem says that for a linear map $T:V\to W$ with finite-dimensional domain of a function $V$,Take a basis $k_1,\dots ,k_r$ of the kernel of a linear map and use basis extension to obtain a basis $k_1,\dots ,k_r,u_1,\dots ,u_s$ of $V$. We show that $T(u_1),\dots ,T(u_s)$ is a basis of the image of a linear map.

Every $v\in V$ is a linear combination of the extended basis, and its $k_i$ terms vanish under $T$. Therefore $T(v)$ is a linear combination of the $T(u_j)$, proving the required linear span property. For linear independence, suppose ${\sum }_j{b}_{j}T(u_j)=0$. Then ${\sum }_j{b}_j{u}_j\in \ker T$, so ${\sum }_j{b}_j{u}_j={\sum }_i{a}_i{k}_i$ for some real scalars $a_i$. The extended basis is linearly independent, forcing every $a_i$ and $b_j$ to be zero. Thus the image of a linear map has dimension $s$, the kernel of a linear map has dimension $r$, and $\dim V=r+s$ proves the result.

A basis provides unique coordinates. DefineThe uniqueness of the coordinates makes this a well-defined function. Adding coordinates and multiplying them by a real scalar shows directly that $\alpha (au+bw)=a\alpha (u)+b\alpha (w)$, so it is a linear map. Substituting $e_i$ gives $\alpha (e_i)=v_i$.

Conversely, every linear map with those three values must send ${\sum }_i{a}_i{e}_i$ to ${\sum }_i{a}_i{v}_i$, by preservation of linear combinations. This determines its value on every element of $V$, proving uniqueness. No linear independence of $v_1,v_2,v_3$ is needed.

For an $m\times n$ matrix $A$ over the real numbers, write its rows as $r_1,\dots ,r_m$ and its columns as $c_1,\dots ,c_n$. ThenThus the row space and column space consist of all linear combinations of the corresponding rows and columns; the null space is the kernel of a linear map represented by $A$. The matrix rank counts independent row or column directions, with row rank equal to column rank.

If $x\in \mathrm{Null}(A)$, the $i$th entry of $Ax=0$ says $r_i\cdot x=0$. Any $y\in \mathrm{Row}(A)$ is a linear combination $y={\sum }_i{\alpha }_i{r}_i$, so bilinearity of the dot product givesHenceThis proves the orthogonality for every pair of vectors in the two vector subspaces. In fact the null space equals the orthogonal complement of the row space, since being orthogonal to every row is exactly the system $Ax=0$.

Write the columns of $A$ as $a_1,\dots ,a_n$ and the coordinates of the vector $x$ as $x_1,\dots ,x_n$. By the definition of matrix multiplication,Indeed, its $i$th coordinate is ${\sum }_{j=1}^n{A}_{ij}{x}_j$, exactly the $i$th coordinate of this linear combination. Since the column space contains every linear combination of the columns, the claim holds for every vector $x$.

Denote the five rows by $r_1,\dots ,r_5$. They satisfyThus $r_2,r_3,r_4$ span the row space. They are linearly independent: in a linear combination $\alpha r_2+\beta r_3+\gamma r_4=0$, coordinate 1 gives $\alpha =0$, coordinate 2 then gives $\beta =0$, and coordinate 4 gives $\gamma =0$. ThereforeAs an alternative basis, the nonzero rows of the reduced row echelon form are

The linearity of the Fibonacci recurrence allows a linear combinationThe initial conditions give $A+B=0$ and $A\varphi +B\psi =1$. Since $\varphi -\psi =\sqrt{5}$, we obtain $A=1/\sqrt{5}$ and $B=-1/\sqrt{5}$. Thus Binet's formula is

Denote the columns by $c_1,\dots ,c_4$. The first two are linearly independent, because the top two coordinates have determinant $3\cdot 2-(-1)\cdot 1=7$. Direct calculation givesThus a column-space basis is $\{(3,1,9{)}^T,(-1,2,4{)}^T\}$, andFor the last description, the third row equals twice the first row plus three times the second, so every column lies in that plane. Since the two independent columns already span a two-dimensional vector subspace, they fill the plane.

The target vector satisfies $5=2(-2)+3(3)$, so it is in the column space. An especially simple linear combination isIn terms of the displayed basis, the same vector is $-c_1/7+11c_2/7$. The column space itself is unique; the freedom is in choosing its basis.

The vectors $v_1,\dots ,v_n$ are linearly independent ifThey are linearly dependent if some such relation has at least one nonzero coefficient. All coefficients lie in the underlying field. Thus linear independence means that only the trivial linear combination gives zero; linear dependence means that a nontrivial one does.